package com.seu.data.structures.setAndMap.leetcodeSet;

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

// Leetcode 804. Unique Morse Code Words
// https://leetcode.com/problems/unique-morse-code-words/description/

/**
 * 使用二叉搜索树Set解决 莫尔斯码的独特字数
 *
 * @author liangfeihu
 * @since 2018/12/4 15:00
 */
public class BSTSetSolution {

    private class BST<E extends Comparable<E>> {

        private class Node {
            public E e;
            public Node left, right;

            public Node(E e) {
                this.e = e;
                left = null;
                right = null;
            }
        }

        private Node root;
        private int size;

        public BST() {
            root = null;
            size = 0;
        }

        public int size() {
            return size;
        }

        public boolean isEmpty() {
            return size == 0;
        }

        // 向二分搜索树中添加新的元素e
        public void add(E e) {
            root = add(root, e);
        }

        // 向以node为根的二分搜索树中插入元素e，递归算法
        // 返回插入新节点后二分搜索树的根
        private Node add(Node node, E e) {

            if (node == null) {
                size++;
                return new Node(e);
            }

            if (e.compareTo(node.e) < 0)
                node.left = add(node.left, e);
            else if (e.compareTo(node.e) > 0)
                node.right = add(node.right, e);

            return node;
        }

        // 看二分搜索树中是否包含元素e
        public boolean contains(E e) {
            return contains(root, e);
        }

        // 看以node为根的二分搜索树中是否包含元素e, 递归算法
        private boolean contains(Node node, E e) {

            if (node == null)
                return false;

            if (e.compareTo(node.e) == 0)
                return true;
            else if (e.compareTo(node.e) < 0)
                return contains(node.left, e);
            else // e.compareTo(node.e) > 0
                return contains(node.right, e);
        }

        // 二分搜索树的前序遍历
        public void preOrder() {
            preOrder(root);
        }

        // 前序遍历以node为根的二分搜索树, 递归算法
        private void preOrder(Node node) {

            if (node == null)
                return;

            System.out.println(node.e);
            preOrder(node.left);
            preOrder(node.right);
        }

        // 二分搜索树的非递归前序遍历
        public void preOrderNR() {

            Stack<Node> stack = new Stack<>();
            stack.push(root);
            while (!stack.isEmpty()) {
                Node cur = stack.pop();
                System.out.println(cur.e);

                if (cur.right != null)
                    stack.push(cur.right);
                if (cur.left != null)
                    stack.push(cur.left);
            }
        }

        // 二分搜索树的中序遍历
        public void inOrder() {
            inOrder(root);
        }

        // 中序遍历以node为根的二分搜索树, 递归算法
        private void inOrder(Node node) {

            if (node == null)
                return;

            inOrder(node.left);
            System.out.println(node.e);
            inOrder(node.right);
        }

        // 二分搜索树的后序遍历
        public void postOrder() {
            postOrder(root);
        }

        // 后序遍历以node为根的二分搜索树, 递归算法
        private void postOrder(Node node) {

            if (node == null)
                return;

            postOrder(node.left);
            postOrder(node.right);
            System.out.println(node.e);
        }

        // 二分搜索树的层序遍历
        public void levelOrder() {

            Queue<Node> q = new LinkedList<>();
            q.add(root);
            while (!q.isEmpty()) {
                Node cur = q.remove();
                System.out.println(cur.e);

                if (cur.left != null)
                    q.add(cur.left);
                if (cur.right != null)
                    q.add(cur.right);
            }
        }

        // 寻找二分搜索树的最小元素
        public E minimum() {
            if (size == 0)
                throw new IllegalArgumentException("BST is empty!");

            return minimum(root).e;
        }

        // 返回以node为根的二分搜索树的最小值所在的节点
        private Node minimum(Node node) {
            if (node.left == null)
                return node;
            return minimum(node.left);
        }

        // 寻找二分搜索树的最大元素
        public E maximum() {
            if (size == 0)
                throw new IllegalArgumentException("BST is empty");

            return maximum(root).e;
        }

        // 返回以node为根的二分搜索树的最大值所在的节点
        private Node maximum(Node node) {
            if (node.right == null)
                return node;

            return maximum(node.right);
        }

        // 从二分搜索树中删除最小值所在节点, 返回最小值
        public E removeMin() {
            E ret = minimum();
            root = removeMin(root);
            return ret;
        }

        // 删除掉以node为根的二分搜索树中的最小节点
        // 返回删除节点后新的二分搜索树的根
        private Node removeMin(Node node) {

            if (node.left == null) {
                Node rightNode = node.right;
                node.right = null;
                size--;
                return rightNode;
            }

            node.left = removeMin(node.left);
            return node;
        }

        // 从二分搜索树中删除最大值所在节点
        public E removeMax() {
            E ret = maximum();
            root = removeMax(root);
            return ret;
        }

        // 删除掉以node为根的二分搜索树中的最大节点
        // 返回删除节点后新的二分搜索树的根
        private Node removeMax(Node node) {

            if (node.right == null) {
                Node leftNode = node.left;
                node.left = null;
                size--;
                return leftNode;
            }

            node.right = removeMax(node.right);
            return node;
        }

        // 从二分搜索树中删除元素为e的节点
        public void remove(E e) {
            root = remove(root, e);
        }

        // 删除掉以node为根的二分搜索树中值为e的节点, 递归算法
        // 返回删除节点后新的二分搜索树的根
        private Node remove(Node node, E e) {

            if (node == null)
                return null;

            if (e.compareTo(node.e) < 0) {
                node.left = remove(node.left, e);
                return node;
            } else if (e.compareTo(node.e) > 0) {
                node.right = remove(node.right, e);
                return node;
            } else {   // e.compareTo(node.e) == 0

                // 待删除节点左子树为空的情况
                if (node.left == null) {
                    Node rightNode = node.right;
                    node.right = null;
                    size--;
                    return rightNode;
                }

                // 待删除节点右子树为空的情况
                if (node.right == null) {
                    Node leftNode = node.left;
                    node.left = null;
                    size--;
                    return leftNode;
                }

                // 待删除节点左右子树均不为空的情况

                // 找到比待删除节点大的最小节点, 即待删除节点右子树的最小节点
                // 用这个节点顶替待删除节点的位置
                Node successor = minimum(node.right);
                successor.right = removeMin(node.right);
                successor.left = node.left;

                node.left = node.right = null;

                return successor;
            }
        }

        @Override
        public String toString() {
            StringBuilder res = new StringBuilder();
            generateBSTString(root, 0, res);
            return res.toString();
        }

        // 生成以node为根节点，深度为depth的描述二叉树的字符串
        private void generateBSTString(Node node, int depth, StringBuilder res) {

            if (node == null) {
                res.append(generateDepthString(depth) + "null\n");
                return;
            }

            res.append(generateDepthString(depth) + node.e + "\n");
            generateBSTString(node.left, depth + 1, res);
            generateBSTString(node.right, depth + 1, res);
        }

        private String generateDepthString(int depth) {
            StringBuilder res = new StringBuilder();
            for (int i = 0; i < depth; i++)
                res.append("--");
            return res.toString();
        }
    }

    private interface Set<E> {

        void add(E e);

        boolean contains(E e);

        void remove(E e);

        int getSize();

        boolean isEmpty();
    }

    private class BSTSet<E extends Comparable<E>> implements Set<E> {

        private BST<E> bst;

        public BSTSet() {
            bst = new BST<>();
        }

        @Override
        public int getSize() {
            return bst.size();
        }

        @Override
        public boolean isEmpty() {
            return bst.isEmpty();
        }

        @Override
        public void add(E e) {
            bst.add(e);
        }

        @Override
        public boolean contains(E e) {
            return bst.contains(e);
        }

        @Override
        public void remove(E e) {
            bst.remove(e);
        }
    }

    public int uniqueMorseRepresentations(String[] words) {

        String[] codes = {".-", "-...", "-.-.", "-..", ".", "..-.", "--.", "....", "..", ".---", "-.-", ".-..", "--", "-.", "---", ".--.", "--.-", ".-.", "...", "-", "..-", "...-", ".--", "-..-", "-.--", "--.."};
        BSTSet<String> set = new BSTSet<>();
        for (String word : words) {
            StringBuilder res = new StringBuilder();
            for (int i = 0; i < word.length(); i++)
                res.append(codes[word.charAt(i) - 'a']);

            set.add(res.toString());
        }

        return set.getSize();
    }
}
